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I have a doubt regarding one argument used to prove that the tensor product indeed has the universal property, namely that for any multilinear map $f:V_1\times\cdots\times V_p\to W$ if the tensor product is $T: V_1 \times \cdots \times V_p \to V_1 \otimes \cdots \otimes V_p$ then it's possible to find a unique linear map $h : V_1 \otimes \cdots \otimes V_p \to W$ such that we have $f = h \circ T$.

Well, the author's argument is: let $\mathcal{M}$ be the free vector space on $V_1 \times \cdots \times V_p$ and let $\mathcal{M}_0$ be the set generated by all vectors of the form:

$$(v_1,\dots,v'_i+v''_i,\dots,v_p)-(v_1,\dots,v'_i,\dots,v_p)-(v_1,\dots,v''_i,\dots,v_p)$$

$$(v_1,\dots,av_i,\dots,v_p)-a(v_1,\dots,v_i,\dots,v_p)$$

Then if we define $g : \mathcal{M} \to W$ by it's values on the basis of $\mathcal{M}$ setting the value of $g$ equal to the value of $f$, i,e.: $g(v_1, \dots, v_p) = f(v_1, \dots, v_p)$ then it's pretty clear that by virtue of the multilinearity of $f$ we have $g(\mathcal{M}_0)=0$ and hence $\mathcal{M}_0\subset \ker g$. After that the author says: "because of that, $g$ induces the map $h$".

He says that it's because of one property that says that if $f : L \to M$ and $g : L \to N$ are such that $\ker f \subset \ker g$ then there's $h : N \to M$ such that $g = f\circ h$, however I cannot see where this proposition fits into this case.

Can someone clarify the argument of the author ?

Thanks in advance for the help.

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1  
I would think $g = f \circ f$ needs some correction? What does that have to do with $h$? –  James S. Cook Apr 2 '13 at 23:39
    
Sorry, I wrote wrong and didn't notice. I'll correct. –  user1620696 Apr 2 '13 at 23:42

2 Answers 2

up vote 2 down vote accepted

You have $g : \mathcal{M} \to W$ and $\pi : \mathcal{M} \to \mathcal{M} / \mathcal{M}_0$ (with the intention to set $V_1 \otimes \cdots \otimes V_p := \mathcal{M} / \mathcal{M}_0$).

As $\ker \pi = \mathcal{M}_0 \subseteq \ker g$ there is $h: \mathcal{M} / \mathcal{M}_0 \to W$ with $g = h \circ \pi$.

That should answer your question, to complete the construction you may need the following:

If $\iota: V_1\times\cdots\times V_p \to \mathcal{M}$ was the obvious embedding, set $T := \pi \circ \iota$. You need to show that $T$ is multilinear, using the properties of $\mathcal{M}_0$.

As $\iota(V_1\times\cdots\times V_p)$ spans $\mathcal{M}$, $T(V_1\times\cdots\times V_p)$ spans $\mathcal{M}/\mathcal{M}_0$, and therefore only one $h$ exists with $f = h \circ T$.

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If we have an equivalence relation $\sim$ on a set $A$, and consider the quotient set $A/\sim$, then a function $f:A\to B$ factors through the canonical surjection $A\to A/\sim$ if and only if $\forall x,y\in A:\, x\sim y\Rightarrow f(x)=f(y)$.

In the case of (Abelian) groups or vector spaces, the structure preserving equivalence relations $\sim$ correspond to (normal) subgroups / subspaces $H$, by defining $$a\sim b \iff a-b\in H\,.$$

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